Prove that a cyclic group with only one generator can have at most 2 elements












12












$begingroup$



Prove that a cyclic group that has only one generator has at most $2$ elements.




I want to know if my proof would be valid:



Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






abstract-algebra greatest-common-divisor cyclic-groups







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    12












    $begingroup$



    Prove that a cyclic group that has only one generator has at most $2$ elements.




    I want to know if my proof would be valid:



    Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



    I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






    abstract-algebra greatest-common-divisor cyclic-groups







    share|cite|improve this question









    New contributor




    Pablo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      12












      12








      12





      $begingroup$



      Prove that a cyclic group that has only one generator has at most $2$ elements.




      I want to know if my proof would be valid:



      Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



      I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






      abstract-algebra greatest-common-divisor cyclic-groups







      share|cite|improve this question









      New contributor




      Pablo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$





      Prove that a cyclic group that has only one generator has at most $2$ elements.




      I want to know if my proof would be valid:



      Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



      I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






      abstract-algebra greatest-common-divisor cyclic-groups




      abstract-algebra greatest-common-divisor cyclic-groups






      share|cite|improve this question









      New contributor




      Pablo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question









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      share|cite|improve this question








      edited 15 hours ago









      darij grinberg

      11.3k33164




      11.3k33164






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      asked 16 hours ago









      PabloPablo

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      New contributor





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          3 Answers
          3






          active

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          16












          $begingroup$

          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            8 hours ago






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            3 hours ago





















          9












          $begingroup$

          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            8 hours ago



















          0












          $begingroup$

          Here is another take.



          The number of generators is $phi(n)$, where $phi$ is Euler's function.



          Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



          In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



          In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



          Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



          (The key fact is: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            The argument in the very last sentence is wrong (though the claim is true).
            $endgroup$
            – darij grinberg
            14 hours ago












          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            5 hours ago











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          3 Answers
          3






          active

          oldest

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          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          16












          $begingroup$

          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            8 hours ago






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            3 hours ago


















          16












          $begingroup$

          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            8 hours ago






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            3 hours ago
















          16












          16








          16





          $begingroup$

          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






          share|cite|improve this answer









          $endgroup$



          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 16 hours ago









          MPWMPW

          30.2k12157




          30.2k12157












          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            8 hours ago






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            3 hours ago




















          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            8 hours ago






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            3 hours ago


















          $begingroup$
          Beautiful! Thank you so much for this!
          $endgroup$
          – Pablo
          8 hours ago




          $begingroup$
          Beautiful! Thank you so much for this!
          $endgroup$
          – Pablo
          8 hours ago




          2




          2




          $begingroup$
          When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
          $endgroup$
          – Marc van Leeuwen
          3 hours ago






          $begingroup$
          When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
          $endgroup$
          – Marc van Leeuwen
          3 hours ago













          9












          $begingroup$

          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            8 hours ago
















          9












          $begingroup$

          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            8 hours ago














          9












          9








          9





          $begingroup$

          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






          share|cite|improve this answer









          $endgroup$



          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 15 hours ago









          LBJFSLBJFS

          2307




          2307












          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            8 hours ago


















          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            8 hours ago
















          $begingroup$
          That's a very good observation, thanks!
          $endgroup$
          – Pablo
          8 hours ago




          $begingroup$
          That's a very good observation, thanks!
          $endgroup$
          – Pablo
          8 hours ago











          0












          $begingroup$

          Here is another take.



          The number of generators is $phi(n)$, where $phi$ is Euler's function.



          Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



          In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



          In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



          Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



          (The key fact is: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            The argument in the very last sentence is wrong (though the claim is true).
            $endgroup$
            – darij grinberg
            14 hours ago












          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            5 hours ago
















          0












          $begingroup$

          Here is another take.



          The number of generators is $phi(n)$, where $phi$ is Euler's function.



          Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



          In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



          In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



          Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



          (The key fact is: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            The argument in the very last sentence is wrong (though the claim is true).
            $endgroup$
            – darij grinberg
            14 hours ago












          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            5 hours ago














          0












          0








          0





          $begingroup$

          Here is another take.



          The number of generators is $phi(n)$, where $phi$ is Euler's function.



          Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



          In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



          In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



          Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



          (The key fact is: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$



          Here is another take.



          The number of generators is $phi(n)$, where $phi$ is Euler's function.



          Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



          In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



          In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



          Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



          (The key fact is: if $d$ divides $n$, then $phi(n) ge phi(d)$.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 5 hours ago

























          answered 15 hours ago









          lhflhf

          165k10171396




          165k10171396








          • 1




            $begingroup$
            The argument in the very last sentence is wrong (though the claim is true).
            $endgroup$
            – darij grinberg
            14 hours ago












          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            5 hours ago














          • 1




            $begingroup$
            The argument in the very last sentence is wrong (though the claim is true).
            $endgroup$
            – darij grinberg
            14 hours ago












          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            5 hours ago








          1




          1




          $begingroup$
          The argument in the very last sentence is wrong (though the claim is true).
          $endgroup$
          – darij grinberg
          14 hours ago






          $begingroup$
          The argument in the very last sentence is wrong (though the claim is true).
          $endgroup$
          – darij grinberg
          14 hours ago














          $begingroup$
          @darij, thanks.
          $endgroup$
          – lhf
          5 hours ago




          $begingroup$
          @darij, thanks.
          $endgroup$
          – lhf
          5 hours ago










          Pablo is a new contributor. Be nice, and check out our Code of Conduct.










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          Pablo is a new contributor. Be nice, and check out our Code of Conduct.













          Pablo is a new contributor. Be nice, and check out our Code of Conduct.












          Pablo is a new contributor. Be nice, and check out our Code of Conduct.
















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